Question #203419

Given that š‘“(š‘„) = 3š‘„ 2 āˆ’ 4š‘„ + 7, use the definition of the derivative to find š‘“ ′ (š‘„).


Expert's answer

Given that š‘“(š‘„)=3š‘„2āˆ’4š‘„+7š‘“(š‘„) = 3š‘„^2 āˆ’ 4š‘„ + 7 , let us find š‘“ā€²(š‘„)š‘“'(š‘„):


f′(x)=lim⁔Δx→0f(x+Ī”x)āˆ’f(x)Ī”x=lim⁔Δx→03(x+Ī”x)2āˆ’4(x+Ī”x)+7āˆ’(3x3āˆ’4x+7)Ī”x=lim⁔Δx→03x3+6xĪ”x+3(Ī”x)2āˆ’4xāˆ’4Ī”x+7āˆ’3x2+4xāˆ’7Ī”x=lim⁔Δx→06xĪ”x+3(Ī”x)2āˆ’4Ī”xĪ”x=lim⁔Δx→0(6x+3Ī”xāˆ’4)=6xāˆ’4.f'(x)=\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}= \lim\limits_{\Delta x\to 0}\frac{3(x+\Delta x)^2-4(x+\Delta x)+7-(3x^3-4x+7)}{\Delta x}= \lim\limits_{\Delta x\to 0}\frac{3x^3+6x\Delta x+3(\Delta x)^2-4x-4\Delta x+7-3x^2+4x-7}{\Delta x}= \lim\limits_{\Delta x\to 0}\frac{6x\Delta x+3(\Delta x)^2-4\Delta x}{\Delta x}= \lim\limits_{\Delta x\to 0}(6x+3\Delta x-4)=6x-4.



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